If f(x) = sin^{-1}(sin x); x in R,then f is | vedclass

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(B) The function $f(x) = \sin^{-1}(\sin x)$ is a periodic function with period $2\pi$. It is defined as: $f(x) = \begin{cases} x - 2k\pi, & 2k\pi - \f

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continuous for all $x$ but not differentiable for all $x = (2k + 1)\frac{\pi}{2}, k \in I$

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(B) The function $f(x) = \sin^{-1}(\sin x)$ is a periodic function with period $2\pi$. It is defined as: $f(x) = \begin{cases} x - 2k\pi, & 2k\pi - \frac{\pi}{2} \le x \le 2k\pi + \frac{\pi}{2} \ (2k+1)\pi - x, & 2k\pi + \frac{\pi}{2} where $k \in I$. Since $\sin^{-1}(\sin x)$ is a continuous function for all $x \in R$,it is continuous everywhere. However,the function is not differentiable at points where the slope changes abruptly,which are the points $x = (2k + 1)\frac{\pi}{2}$ for all $k \in I$. Thus,the function is continuous for all $x$ but not differentiable at $x = (2k + 1)\frac{\pi}{2}$.

If $f(x) = \sin^{-1}(\sin x)$; $x \in R$,then $f$ is

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